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In my HiFi system the CD player is playing way too loud compared to the other sources (radio, streaming gadgets, tv. etc)

I figure a simple voltage divider say 33kohm and 27kohm would reduce with about 6dB. Simply hack the circuit into the cd player, where there is plenty physical room around the connector inside.

Something like this

*---,
    |
   33K       
    |
    +-----*     
    |
   27K
    |
*---+-----*

But, will this hack be bad for the fidelity?

Should I aim for 50kohm as imput impedance or is best practice different when we are dealing with stereos from '90?

Also, a friend of mine told me to add a 220nF cap in series with the output to block DC components, but I failed to understand where they should come from.

*---,
    |
   33K       
    |    220nF
    +-----||----+----*
    |           |
   27K         56K
    |           |
*---+-----------+----*

The CD is a NAD 5240 and the preamp is a NAD 1155 in case any of you think it might matter.

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2 Answers 2

up vote 3 down vote accepted

If the connection already works (just too loud) there isn't enough DC component to worry about.

The only things it might do to damage fidelity are both minor:

  1. The resistors act as an additional noise source: with the CD paused and the volume fully up you might hear a slight increase in the faint background hiss : by all means test that, but I doubt it. (Return the volume to normal afterwards!)
  2. Slight reduction in high frequency response. As you are putting it in the CD player, the two resistors in parallel (call it 15k) and the cable connecting it to the amp (probably below 1000pf) will roll off the high end slightly.

For C=1000pf, RC = 15us, f=1/(2*pi*RC)=10kHz.

So if the cable (and any input capacitance in the amplifier) came to 1nf, response would be -3dB at 10kHz. This is worth fixing.

  1. You can choose a low capacitance cable and impress your friends hey, cables really DO sound different!!!
  2. You can build the attenuator between socket and plug, and wire it after the cable (so the cable is ahead of the attenuator)
  3. you can reduce both resistors by a factor of 5-10. Say, 6k8 and 5k6, -3dB at 50kHz, -1dB at 25kHz, probably a reasonable compromise
  4. you could build the attenuator into the amplifier instead.
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All makes sense here. I think I should have opted for lower values like you suggest in #3. I don't know the output impedance of the CD player, might be 1K, and you are suggesting loading it with 12K, wouldn't I then be driving it too hard, exposing capacitance in said output? –  mogul Mar 30 at 19:48
    
(3) would be my preferred option too. Zout is unlikely to be as high as 1k but even if it is, you most likely just get 1dB more attenuation. Reducing impedances also reduces the resistor noise "problem". Option (1) is primarily for amusement value. (2) - inline attenuators - has the virtue of not modifying the gear. For classic or valuable gear, it would be the best option. –  Brian Drummond Mar 30 at 20:06
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The first attenuator in the question will attenuate approximately 6 dB (as you said) and so will the 2nd attenuator with the 220 nF capacitor. However, neither will pass-thru the DC value from the output where you are hacking the modification.

If the DC value is important to maintain (irrespective of the AC amplitude) then use a 10uF capacitor in series with the 27 kohm in the first circuit.

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So basically what you are saying is that the simple two-resistor divider should be all that I need. (for each channel of cause) Perfect I will find my soldering iron right away! –  mogul Mar 30 at 8:25
    
If it is important to maintain the dc value you will need to put a capacitor in series with the 27k –  Andy aka Mar 30 at 9:30
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